題面在這裏

題意

...見鏈接吧

sol

在此發一篇模擬退火的題解
不得不說luogu的數據真是太良心啦

一句話解釋模擬退火:在一個慢慢縮小的範圍內隨機狀態尋找最優解，當轉移狀態更優時直接接受，當當前狀態更優時以一定概率\((exp(dlt/T))\)接受
具體實現請參照代碼

#include<bits/stdc++.h>
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<iomanip>
#include<cstring>
#include<vector>
 

#include<cstdio>
#include<string>
#include<bitset>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define sqr(x) ((x)*(x))
#define pb push_back
#define RG register
#define il inline
using namespace std;
const int mod=1e9+7
 
;
const int N=1010;
typedef unsigned long long ull;
typedef vector<int>VI;
typedef long long ll;
typedef double dd;
il ll read(){
    RG ll data=0,w=1;RG char ch=getchar();
    while(ch!=‘-‘&&(ch<‘0‘||ch>‘9‘))ch=getchar();
    if(ch==‘-‘)w=-1,ch=getchar();
    while(ch<=‘9‘&&ch>=‘0‘
 
)data=data*10+ch-48,ch=getchar();
    return data*w;
}

int n;dd mn;
struct point{dd x,y,w;}p[N],ans,now;
il dd make(){return rand()%100000/100000.00;}
il dd dis(point a,point b){return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));}
il dd calc(point x){
    RG dd ret=0;
    for(RG int i=1;i<=n;i++)
        ret+=p[i].w*dis(x,p[i]);
    return ret;
}

il void SA(point x){
    RG dd T=1e6;
    while(T>1e-3){//退火過程
        now.x=ans.x+T*(2*make()-1);
        now.y=ans.y+T*(2*make()-1);
        double dlt=calc(ans)-calc(now);
        if(dlt>0||exp(dlt/T)>make())ans=now;
        T*=0.996;
    }
    for(int i=1;i<=50000;++i)//最後在一定範圍內搜尋到局部最優解
    {
        now.x=ans.x+T*(2*make()-1);
        now.y=ans.y+T*(2*make()-1);
        if(calc(ans)-calc(now)>0)ans=now;
    }
}

int main()
{
    srand(time(NULL)+rand());
    n=read();
    for(RG int i=1;i<=n;i++)
        scanf("%lf%lf%lf",&p[i].x,&p[i].y,&p[i].w);
    ans=(point){make(),make(),0};mn=ans.w=calc(ans);
    SA(ans); 
    printf("%.3lf %.3lf\n",ans.x,ans.y);
    return 0;
}

考試騙分，模擬退火，你值得擁有！

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